# 5.4 Comparison tests  (Page 3/7)

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## Limit comparison test

Let ${a}_{n},\phantom{\rule{0.2em}{0ex}}{b}_{n}\ge 0$ for all $n\ge 1.$

1. If $\underset{n\to \infty }{\text{lim}}{a}_{n}\text{/}{b}_{n}=L\ne 0,$ then $\sum _{n=1}^{\infty }{a}_{n}$ and $\sum _{n=1}^{\infty }{b}_{n}$ both converge or both diverge.
2. If $\underset{n\to \infty }{\text{lim}}{a}_{n}\text{/}{b}_{n}=0$ and $\sum _{n=1}^{\infty }{b}_{n}$ converges, then $\sum _{n=1}^{\infty }{a}_{n}$ converges.
3. If $\underset{n\to \infty }{\text{lim}}{a}_{n}\text{/}{b}_{n}=\infty$ and $\sum _{n=1}^{\infty }{b}_{n}$ diverges, then $\sum _{n=1}^{\infty }{a}_{n}$ diverges.

Note that if ${a}_{n}\text{/}{b}_{n}\to 0$ and $\sum _{n=1}^{\infty }{b}_{n}$ diverges, the limit comparison test gives no information. Similarly, if ${a}_{n}\text{/}{b}_{n}\to \infty$ and $\sum _{n=1}^{\infty }{b}_{n}$ converges, the test also provides no information. For example, consider the two series $\sum _{n=1}^{\infty }1\text{/}\sqrt{n}$ and $\sum _{n=1}^{\infty }1\text{/}{n}^{2}.$ These series are both p -series with $p=1\text{/}2$ and $p=2,$ respectively. Since $p=1\text{/}2>1,$ the series $\sum _{n=1}^{\infty }1\text{/}\sqrt{n}$ diverges. On the other hand, since $p=2<1,$ the series $\sum _{n=1}^{\infty }1\text{/}{n}^{2}$ converges. However, suppose we attempted to apply the limit comparison test, using the convergent $p-\text{series}$ $\sum _{n=1}^{\infty }1\text{/}{n}^{3}$ as our comparison series. First, we see that

$\frac{1\text{/}\sqrt{n}}{1\text{/}{n}^{3}}=\frac{{n}^{3}}{\sqrt{n}}={n}^{5\text{/}2}\to \infty \phantom{\rule{0.2em}{0ex}}\text{as}\phantom{\rule{0.2em}{0ex}}n\to \infty .$

Similarly, we see that

$\frac{1\text{/}{n}^{2}}{1\text{/}{n}^{3}}=n\to \infty \phantom{\rule{0.2em}{0ex}}\text{as}\phantom{\rule{0.2em}{0ex}}n\to \infty .$

Therefore, if ${a}_{n}\text{/}{b}_{n}\to \infty$ when $\sum _{n=1}^{\infty }{b}_{n}$ converges, we do not gain any information on the convergence or divergence of $\sum _{n=1}^{\infty }{a}_{n}.$

## Using the limit comparison test

For each of the following series, use the limit comparison test to determine whether the series converges or diverges. If the test does not apply, say so.

1. $\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}+1}$
2. $\sum _{n=1}^{\infty }\frac{{2}^{n}+1}{{3}^{n}}$
3. $\sum _{n=1}^{\infty }\frac{\text{ln}\left(n\right)}{{n}^{2}}$
1. Compare this series to $\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}.$ Calculate
$\underset{n\to \infty }{\text{lim}}\phantom{\rule{0.2em}{0ex}}\frac{1\text{/}\left(\sqrt{n}+1\right)}{1\text{/}\sqrt{n}}=\underset{n\to \infty }{\text{lim}}\begin{array}{c}\frac{\sqrt{n}}{\sqrt{n}+1}\\ \phantom{\rule{0.2em}{0ex}}\end{array}=\underset{n\to \infty }{\text{lim}}\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{/}\sqrt{n}}{1+1\text{/}\sqrt{n}}=1.$
By the limit comparison test, since $\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}$ diverges, then $\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}+1}$ diverges.
2. Compare this series to $\sum _{n=1}^{\infty }{\left(\frac{2}{3}\right)}^{n}.$ We see that
$\underset{n\to \infty }{\text{lim}}\frac{\left({2}^{n}+1\right)\text{/}{3}^{n}}{{2}^{n}\text{/}{3}^{n}}=\underset{n\to \infty }{\text{lim}}\frac{{2}^{n}+1}{{3}^{n}}·\frac{{3}^{n}}{{2}^{n}}=\underset{n\to \infty }{\text{lim}}\frac{{2}^{n}+1}{{2}^{n}}=\underset{n\to \infty }{\text{lim}}\left[1+{\left(\frac{1}{2}\right)}^{n}\right]=1.$

Therefore,
$\underset{n\to \infty }{\text{lim}}\frac{\left({2}^{n}+1\right)\text{/}{3}^{n}}{{2}^{n}\text{/}{3}^{n}}=1.$

Since $\sum _{n=1}^{\infty }{\left(\frac{2}{3}\right)}^{n}$ converges, we conclude that $\sum _{n=1}^{\infty }\frac{{2}^{n}+1}{{3}^{n}}$ converges.
3. Since $\text{ln}\phantom{\rule{0.1em}{0ex}}n compare with $\sum _{n=1}^{\infty }\frac{1}{n}.$ We see that
$\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n\text{/}{n}^{2}}{1\text{/}n}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n}{{n}^{2}}·\frac{n}{1}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n}{n}.$

In order to evaluate $\underset{n\to \infty }{\text{lim}}\text{ln}\phantom{\rule{0.1em}{0ex}}n\text{/}n,$ evaluate the limit as $x\to \infty$ of the real-valued function $\text{ln}\left(x\right)\text{/}x.$ These two limits are equal, and making this change allows us to use L’Hôpital’s rule. We obtain
$\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{x}=\underset{x\to \infty }{\text{lim}}\frac{1}{x}=0.$

Therefore, $\underset{n\to \infty }{\text{lim}}\text{ln}\phantom{\rule{0.1em}{0ex}}n\text{/}n=0,$ and, consequently,
$\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n\text{/}{n}^{2}}{1\text{/}n}=0.$

Since the limit is $0$ but $\sum _{n=1}^{\infty }\frac{1}{n}$ diverges, the limit comparison test does not provide any information.
Compare with $\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}$ instead. In this case,
$\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.2em}{0ex}}n\text{/}{n}^{2}}{1\text{/}{n}^{2}}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.2em}{0ex}}n}{{n}^{2}}·\frac{{n}^{2}}{1}=\underset{n\to \infty }{\text{lim}}\text{ln}\phantom{\rule{0.1em}{0ex}}n=\infty .$

Since the limit is $\infty$ but $\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}$ converges, the test still does not provide any information.
So now we try a series between the two we already tried. Choosing the series $\sum _{n=1}^{\infty }\frac{1}{{n}^{3\text{/}2}},$ we see that
$\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n\text{/}{n}^{2}}{1\text{/}{n}^{3\text{/}2}}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n}{{n}^{2}}·\frac{{n}^{3\text{/}2}}{1}=\underset{n\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n}{\sqrt{n}}.$

As above, in order to evaluate $\underset{n\to \infty }{\text{lim}}\text{ln}\phantom{\rule{0.1em}{0ex}}n\text{/}\sqrt{n},$ evaluate the limit as $x\to \infty$ of the real-valued function $\text{ln}\phantom{\rule{0.1em}{0ex}}x\text{/}\sqrt{x}.$ Using L’Hôpital’s rule,
$\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{\sqrt{x}}=\underset{x\to \infty }{\text{lim}}\frac{2\sqrt{x}}{x}=\underset{x\to \infty }{\text{lim}}\frac{2}{\sqrt{x}}=0.$

Since the limit is $0$ and $\sum _{n=1}^{\infty }\frac{1}{{n}^{3\text{/}2}}$ converges, we can conclude that $\sum _{n=1}^{\infty }\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n}{{n}^{2}}$ converges.

Use the limit comparison test to determine whether the series $\sum _{n=1}^{\infty }\frac{{5}^{n}}{{3}^{n}+2}$ converges or diverges.

The series diverges.

## Key concepts

• The comparison tests are used to determine convergence or divergence of series with positive terms.
• When using the comparison tests, a series $\sum _{n=1}^{\infty }{a}_{n}$ is often compared to a geometric or p -series.

Use the comparison test to determine whether the following series converge.

$\sum _{n=1}^{\infty }{a}_{n}$ where ${a}_{n}=\frac{2}{n\left(n+1\right)}$

$\sum _{n=1}^{\infty }{a}_{n}$ where ${a}_{n}=\frac{1}{n\left(n+1\text{/}2\right)}$

Converges by comparison with $1\text{/}{n}^{2}.$

#### Questions & Answers

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Damian Reply
research.net
kanaga
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
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Anassong
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s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
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Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
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Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
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s. Reply
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Harper
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s.
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SUYASH Reply
for screen printed electrodes ?
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s. Reply
of graphene you mean?
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or in general
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in general
s.
Graphene has a hexagonal structure
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Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
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You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?
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